Integral of rational function $\int\frac{x^2-18x-1}{(x-1)^2(x^2+1)} \text{d}x$

Question

Evaluate $$\int\dfrac{x^2-18x-1}{(x-1)^2(x^2+1)} \text{d}x$$ (Remember to use $\ln |u|$ where appropriate. Use $\text{C}$ for the constant of integration.)

I got the answer $$\dfrac{-1}{2 \ln(x^2+1)} + \dfrac{9}{x+1+\ln|x-1|}+9\tan^{-1}(x) + \text{C}.$$ Is this correct?

Thanks everyone for the help i got the answer!

Answer 1

Yes, it's correct. Just don't forget the absolute values (replace $\ln(x - 1)$ with $\ln|x - 1|$).